 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Rhomboctagon (Posted on 2013-09-28) ABCD is a rhombus with side length 13. Equilateral triangles are erected on all four sides resulting in a concave octagon APBQCRDS. Given PQ=24, compute the area of the octagon.

 No Solution Yet Submitted by Danish Ahmed Khan No Rating Comments: ( Back to comment list | You must be logged in to post comments.) Solutions Comment 1 of 1
First, there are two possibilities:  Segment PQ can lie either (case 1) inside or (case 2) outside of the rhombus.

In both cases the area of the octagon is the area of the rhombus plus four equilateral triangles with side of 13 and so total area of 338sqrt(3).  So all we really need is the area of the rhombus and we can add this.

Call the center of the rhombus O.

Case 1: extend AC to meet QR at T.  Triangle CTQ is a 5-12-13 right triangle so angle QCT=arctan(12/5)
Since angle BCQ=60, angle OCB=120-arctan(12/5)

Now find OC=13cos(120-arctan(12/5))=(-5+12sqrt(3))/2
And OB=13sin(120-arctan(12/5))=(12+5sqrt(3))/2
The area of the rhombus is 2*OC*OB=(120+119sqrt(3))/2
and adding in the equilateral triangles gives a final answer of (120+457sqrt(3))/2

Case 2 is very similar the final solution here is (-120+457sqrt(3))/2

 Posted by Jer on 2013-09-28 15:44:39 Please log in:

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